Rudin's functional analysis theorem 1.21(b) This theorem proves that any finite-dimensional subspace $Y$ of a topological vector space $X$ is closed.
take $p\in \overline Y$ and we know $p\in tV$ for some $t>0$. $V$ is a balanced neighborhood of $0$ in $X$. I am confused why $p\in\overline{Y\cap (tV)}$?
In general, $\overline A\cap B\subseteq \overline{A\cap B}$? I don't think it is true, for example:
$A=(0,1),B=[1,2]$ 
 A: If $p\notin \overline{Y\cap (tV)}$, then there is a neighborhood $U_p$ of $p$ such that 
$$(U_p\cap (tV))\cap Y=U_p\cap (Y\cap (tV))\subseteq U_p\cap \overline{Y\cap (tV)}=\emptyset,$$ 
which is an immediate contradiction: $U_p\cap (tV)$ is a neighborhood of $p$ so it must intersect $Y$.
A: I'm aware this is an old question, and also that the OP has already been given a satisfying answer; nonetheless, I've just stumbled upon it, and I think there may be a (minor?) point deserving some closer attention (sorry in advance if my remark is stupid, if so please just delete it).
In a topological space $X$, for $Y\subset X$, $p\in X$ and $U$ a neighborhood of $p$, we have:
$$p\in \overline{Y}\implies p\in\overline{Y\cap U}.$$
In fact, if $V$ is a neighborhood of $p$, $V\cap U$ is a neighborhood of $p$ as well, so:
$$(Y\cap U)\cap V\ne \varnothing.$$
[As a consequence, if $G$ is an open set: $\overline{Y} \cap G\subset \overline{Y\cap G}$.]
If the result $p\in\overline{Y\cap tV}$ is to be obtained by following the above line of reasoning, we'd need $tV$ to be a neighborhood of $p$, i.e. $p$ to be not just an object of $tV$, but an object of its interior.
If Rudin here is assuming only that $t>0$ is big enough so that $p\in tV$, either he is following a different (and I guess better) line of reasoning I could not think of, or he forgot to make some (stronger but unproblematic!) assumption explicit (here it would suffice to assume that $V$ is an open neighborhood, or that $t>0$ is big enough to have $p$ in the interior of $tV$) — or is he just using some uncommon convention?, such as that a neighborhood is an open set?
